ON A CONJECTURE ON SHIFTED PRIMES WITH LARGE PRIME FACTORS, II

Abstract

Let $\mathcal {P}$ be the set of primes and $\pi (x)$ the number of primes not exceeding x. Let $P^+(n)$ be the largest prime factor of n, with the convention $P^+(1)=1$, and $ T_c(x)=\#\{p\le x:p\in \mathcal {P},P^+(p-1)\ge p^c\}. $ Motivated by a conjecture of Chen and Chen [‘On the largest prime factor of shifted primes’, Acta Math. Sin. (Engl. Ser.) 33 (2017), 377–382], we show that for any c with $8/9\le c<1$,

$$ \begin{align*} \limsup_{x\rightarrow\infty}T_c(x)/\pi(x)\le 8(1/c-1), \end{align*} $$

which clearly means that

$$ \begin{align*} \limsup_{x\rightarrow\infty}T_c(x)/\pi(x)\rightarrow 0 \quad \text{as } c\rightarrow 1. \end{align*} $$

1. Introduction

The investigation of shifted primes with large prime factors began in a brilliant article of Goldfeld [12]. Historically, this topic had aroused great interest because of its unexpected connection with the first case of Fermat’s last theorem, thanks to the theorems of Fouvry [11] and Adleman and Heath-Brown [1].

For any positive integer n, let $P^+(n)$ be the largest prime factor of n with the convention $P^+(1)=1$ . Let $\mathcal {P}$ be the set of primes and $\pi (x)$ the number of primes not exceeding x. For $0<c<1$ , let $T_c(x)=\#\{p\le x:p\in \mathcal {P},P^+(p-1)\ge p^c\}.$ As early as 1969, Goldfeld [12] proved

$$ \begin{align*} \liminf_{x\rightarrow\infty}T_{{1}/{2}}(x)/\pi(x)\ge 1/2. \end{align*} $$

Goldfeld further remarked that his argument also leads to

(1.1) $$ \begin{align} \liminf_{x\rightarrow\infty}T_{c}(x)/\pi(x)>0, \end{align} $$

provided that $c<{7}/{12}$ . It turns out that exploring large c which satisfy (1.1) is rather difficult and important. For improvements on the values of c, see Motohashi [21], Hooley [15, 16], Deshouillers and Iwaniec [7], and Fouvry [11]. Up to now, the best numerical value of c satisfying (1.1), with a cost of replacing $\pi (x)$ with $\pi (x)/\log x$ , is 0.677, obtained by Baker and Harman [3].

In an earlier note [8] on this topic, I showed that

(1.2) $$ \begin{align} \limsup_{x\rightarrow\infty}T_{c}(x)/\pi(x)<1/2 \end{align} $$

holds for some absolute constant $c<1$ . As a corollary, I disproved a 2017 conjecture of Chen and Chen [6] that

$$ \begin{align*} \liminf_{x\rightarrow\infty}T_{c}(x)/\pi(x)\ge 1/2 \end{align*} $$

for any c with $1/2\le c<1$ . The proof in my earlier note is based on the following deep result which is a corollary of the Brun–Titchmarsh inequality.

Proposition 1.1 [24, Lemma 2.2].

There exist two functions $K_2(\theta )>K_1(\theta )>0$ , defined on the interval $(0,17/32)$ such that for each fixed real $A>0$ and all sufficiently large $Q=x^\theta $ , the inequalities

$$ \begin{align*}K_1(\theta)\frac{\pi(x)}{\varphi(m)}\le \pi(x;m,1)\le K_2(\theta)\frac{\pi(x)}{\varphi(m)}\end{align*} $$

hold for all integers $m\in (Q,2Q]$ with at most $O(Q(\log Q)^{-A})$ exceptions, where the implied constant depends only on A and $\theta $ . Moreover, for any fixed $\varepsilon>0$ , these functions can be chosen to satisfy the following properties:

• • $K_1(\theta )$ is monotonic decreasing and $K_2(\theta )$ is monotonic increasing;

• • $K_1(1/2)=1-\varepsilon $ and $K_2(1/2)=1+\varepsilon $ .

The constant c in (1.2) is not specified because of the indeterminate nature of $K_1(\theta )$ in Proposition 1.1. In fact, $K_1(\theta )$ (and hence c) can be explicitly given if one checks carefully the articles of Baker and Harman [2] for $1/2\le \theta \le 13/25$ , and Mikawa [19] for $13/25\le \theta \le 17/32$ . This gives $K_1(\theta )\ge 0.16$ for $1/2\le \theta \le 13/25$ [2, Theorem 1] and $K_1(\theta )\ge 1/100$ for Mikawa’s range [19, (4)]. However, it seems that the constant c in (1.2) obtained in this way will be very close to $1$ (see the proofs in [8]).

In [8], I also pointed out that Chen and Chen’s conjecture is already in contradiction with the Elliott–Halberstam conjecture (from Pomerance [22], Granville [13], Wang [23] and Wu [24]). In fact,

(1.3) $$ \begin{align} \limsup_{x\rightarrow\infty}T_c(x)/\pi(x)=\lim_{x\rightarrow\infty}T_c(x)/\pi(x) =\bigg(1-\rho\bigg(\frac{1}{c}\bigg)\bigg)\rightarrow 0 \quad \text{as } c\rightarrow 1, \end{align} $$

under the assumption of the Elliott–Halberstam conjecture, where $\rho (u)$ is the Dickman function, defined as the unique continuous solution of the differential-difference equation

$$ \begin{align*} \begin{cases} \rho(u)=1 & \mbox{for }0\le u\le 1, \\ u\rho'(u)=-\rho(u-1) & \mbox{for }u>1. \end{cases} \end{align*} $$

However, there are earlier results related to the conjecture of Chen and Chen, and my earlier result (1.2). In fact, as indicated by the proof of a result of Erdős [9, Lemma 4], as early as 1935, one could already conclude from Erdős’ proof combined with Lemma 2.2 of Wu (see below) that (1.3) is true in part.

Theorem 1.2 (Erdős).

Unconditionally,

$$ \begin{align*} \limsup_{x\rightarrow\infty}T_c(x)/\pi(x)\rightarrow 0 \quad \text{as } c\rightarrow1. \end{align*} $$

Essentially, Theorem 1.2 can be deduced from Erdős’ proof by adding Wu’s lemma (see Erdős’ argument in [9, from page 212, line 6 to page 213, line 4]). Since Erdős’ conclusion is not clearly formulated, it is meaningful to restate it explicitly as Theorem 1.2. It is also of interest to pursue Erdős’ theorem a little further to reach the following quantitative form.

Theorem 1.3. For $8/9\le c<1$ ,

$$ \begin{align*} \limsup_{x\rightarrow\infty}T_c(x)/\pi(x)\le 8(1/c-1). \end{align*} $$

We note that the restriction on $c\ge 8/9$ in our theorem is natural since otherwise, the upper bound would exceed $1$ which is certainly meaningless. Theorem 1.3 can also be compared with the results of Goldfeld [12], Luca et al. [18], and Chen and Chen [6] which state that

$$ \begin{align*} \liminf_{x\rightarrow\infty}T_c(x)/\pi(x)\ge 1-c \end{align*} $$

for $0<c\le 1/2$ . These bounds were recently improved in part by Feng and Wu [10], and Liu, Wu and Xi [17]. From Theorem 1.3, we clearly have two corollaries, one of which is Erdős’ theorem (Theorem 1.2) while the other revisits the main result (1.2) of my earlier note in a quantitative form.

Corollary 1.4. For $c>16/17$ ,

$$ \begin{align*} \limsup_{x\rightarrow\infty}T_{c}(x)/\pi(x)<1/2. \end{align*} $$

2. Proofs

From now on, p will always be a prime. The proof of Theorem 1.3 is based on the following lemma deduced from the sieve method (see, for example, [14, Theorem 5.7, page 172]).

Lemma 2.1. Let g be a natural number and let $a_i,b_i~(i=1,2,\ldots ,g)$ be integers satisfying

$$ \begin{align*} E:=\prod_{i=1}^{g}a_i\prod_{1\le r<s\le g}(a_rb_s-a_sb_r)\neq0. \end{align*} $$

Let $\rho (p)$ denote the number of solutions in n modulo p of

$$ \begin{align*} \prod_{i=1}^{g}(a_in+b_i)\equiv 0{\ (\mathrm{mod}\ p)}, \end{align*} $$

and suppose that

$$ \begin{align*} \rho(p)<p \quad \text{for all } p. \end{align*} $$

If the real numbers y and z satisfy $1<y\le z,$ then

$$ \begin{align*} &\hspace{-13pt}|\{n:z-y<n\le z, a_in+b_i \text{ prime for } i=1,2,\ldots,g\}|\\ &\le 2^gg!\prod_{p}\bigg(1-\frac{\rho(p)-1}{p-1}\bigg)\bigg(1-\frac{1}{p}\bigg)^{-g+1}\frac{y}{\log ^gy} \bigg(1+O\bigg(\frac{\log\log3y+\log\log 3|E|}{\log y}\bigg)\bigg), \end{align*} $$

where the constant implied by the O-symbol depends at most on g.

We also need the following important relation established by Wu [24, Theorem 2].

Lemma 2.2. For $0<c<1$ , let

$$ \begin{align*} T^{\prime}_c(x)=\#\{p\le x:p\in\mathcal{P}, P^+(p-1)\ge x^c\}. \end{align*} $$

Then for sufficiently large x,

$$ \begin{align*} T_c(x)=T^{\prime}_c(x)+O\bigg(\frac{x\log\log x}{(\log x)^2}\bigg). \end{align*} $$

We now turn to the proof of Theorem 1.3.

Proof of Theorem 1.3.

Let x be a sufficiently large number throughout the proof. Instead of investigating $T_c(x)$ , we first deal with $T^{\prime }_c(x)$ . For $1/2\le c<1$ , it is easy to see that

(2.1) $$ \begin{align} T^{\prime}_c(x)= \sum_{\substack{x^c\le q<x\\ q\in \mathcal{P}}}\sum_{\substack{p\le x\\q|p-1}}1. \end{align} $$

On putting $p-1=qh$ in the sum (2.1) and then exchanging the order of summation,

(2.2) $$ \begin{align} T^{\prime}_c(x)=\sum_{\substack{x^c\le q<x \\ q\in \mathcal{P}}}\sum_{\substack{h< x/q\\qh+1\in \mathcal{P}}}1\le \sum_{\substack{h<x^{1-c}\\ 2|h}}\sum_{\substack{2<q<x/h \\ q,qh+1\in \mathcal{P}}}1. \end{align} $$

For any h with $2 \mid h$ and $h<x^{1-c}$ , let $\rho (p)$ denote the number of solutions of

$$ \begin{align*} n(hn+1)\equiv 0{\ (\mathrm{mod}\ p)}. \end{align*} $$

Then

$$ \begin{align*} \rho(p)= \begin{cases} 1 & \text{if } p \mid h, \\ 2 & \text{otherwise.} \end{cases} \end{align*} $$

Now, by Lemma 2.1 with $g=2, a_1=1, b_1=0, a_2=h,b_2=1$ and $z=y=x/h$ ,

$$ \begin{align*} 3|E|=3h\ll x, \quad 3y=3x/h\ll x \quad \text{and} \quad y=x/h\ge \sqrt{x}, \end{align*} $$

from which it follows that

(2.3) $$ \begin{align} \sum_{\substack{2<q<x/h \\ q,qh+1\in \mathcal{P}}}1\le 16\mathfrak{S}\prod_{\substack{p|h\\ p>2}}\bigg(1+\frac{1}{p-2}\bigg) \frac{x/h}{\log^2(x/h)} \bigg(1+O\bigg(\frac{\log\log x}{\log x}\bigg)\bigg), \end{align} $$

where an empty product for $\prod _{p|h, p>2}$ above denotes $1$ as usual and

$$ \begin{align*} \mathfrak{S}=\prod_{p>2}\bigg(1-\frac{1}{p-1}\bigg)\bigg(1-\frac{1}{p}\bigg)^{-1}= \prod_{p>2}\bigg(1-\frac{1}{(p-1)^2}\bigg). \end{align*} $$

Inserting (2.3) into (2.2) gives

(2.4) $$ \begin{align} T^{\prime}_c(x)&\le (1+o(1))16\mathfrak{S}\sum_{\substack{h<x^{1-c}\\ 2|h}}\prod_{\substack{p|h\\ p>2}}\bigg(1+\frac{1}{p-2}\bigg) \frac{x/h}{\log^2(x/h)}. \end{align} $$

Note that

(2.5) $$ \begin{align} \prod_{\substack{p|h\\ p>2}}\bigg(1+\frac{1}{p-2}\bigg)\le 2\prod_{\substack{p|h\\ p>2}}\bigg(1+\frac{1}{p}\bigg) \end{align} $$

since the gaps between odd primes are at least $2$ , from which we can already give a nontrivial upper bound of $T^{\prime }_c(x)$ via partial summations. To make our bound more explicit than (2.5), we employ a nice result of Banks and Shparlinski [4, Lemma 2.3] (on taking $a=1$ therein), which states that for $z\ge 2$ ,

(2.6) $$ \begin{align} S(z):=\sum_{\substack{h<z\\ 2|h}}\frac{1}{h}\prod_{\substack{p|h\\ p>2}}\bigg(1+\frac{1}{p-2}\bigg)=\frac{1+o(1)}{2\mathfrak{S}}\log z. \end{align} $$

For $1\le z<2$ , we set $S(z)=0$ . By partial summation,

(2.7) $$ \begin{align} \sum_{\substack{h<x^{1-c}\\ 2|h}}\prod_{\substack{p|h\\ p>2}}\bigg(1+\frac{1}{p}\bigg)\frac{1/h}{\log^2(x/h)}=\frac{S(x^{1-c})}{(\log x^c)^2}-\int_{1}^{x^{1-c}}S(z)\,d\bigg(\log \frac{x}{z}\bigg)^{-2}. \end{align} $$

Note also that for $z\ge 2$ ,

$$ \begin{align*} S(z)\le \sum_{\substack{h<z}}\frac{1}{h}\prod_{\substack{p|h\\ p>2}}\bigg(1+\frac{1}{p-2}\bigg)\le \sum_{\substack{h<z}}\frac{1}{h}\prod_{\substack{p|h}}\bigg(1+\frac{3}{p}\bigg)=\sum_{\substack{h<z}}\frac{1}{h}\sum_{d|h} \frac{3^{\omega(d)}\mu^2(d)}{d}, \end{align*} $$

where $\mu (d)$ is the Möbius function and $\omega (d)$ is the number of distinct prime factors of d. Exchanging the order of summation,

(2.8) $$ \begin{align} S(z)\le \sum_{d<z}\frac{3^{\omega(d)}\mu^2(d)}{d}\sum_{\substack{h<z\\ d|h}}\frac{1}{h}\le 3\sum_{d<z}\frac{3^{\omega(d)}\mu^2(d)}{d^2}\log z<3K\log z, \end{align} $$

where

$$ \begin{align*} K=3\sum_{d=1}^{\infty}\frac{3^{\omega(d)}\mu^2(d)}{d^2}. \end{align*} $$

From (2.8),

(2.9) $$ \begin{align} \int_{1}^{\log x}S(z)\,d\bigg(\log \frac{x}{z}\bigg)^{-2}\ll_K \frac{\log\log x}{(\log x)^2}=o((\log x)^{-1}). \end{align} $$

Now, routine computations yield

(2.10) $$ \begin{align} \frac{S(x^{1-c})}{(\log x^c)^2}=\frac{1+o(1)}{\mathfrak{S}}\frac{(1-c)}{2c^2}(\log x)^{-1} \end{align} $$

and

(2.11) $$ \begin{align} \int_{\log x}^{x^{1-c}}S(z)\,d\bigg(\log \frac{x}{z}\bigg)^{-2}&=\frac{1+o(1)}{\mathfrak{S}}\int_{\log x}^{x^{1-c}}\frac{\log z}{z}\bigg(\log \frac{x}{z}\bigg)^{-3}\,dz\nonumber\\ &= \frac{1+o(1)}{\mathfrak{S}}\int_{1}^{x^{1-c}}\frac{\log z}{z}\bigg(\log \frac{x}{z}\bigg)^{-3}\,dz+o((\log x)^{-1})\nonumber\\ &=\frac{1+o(1)}{\mathfrak{S}}\int_{x^c}^{x}\frac{\log x-\log u}{u}(\log u)^{-3}\,du+o((\log x)^{-1})\nonumber\\ &=\frac{1+o(1)}{\mathfrak{S}}\bigg(\frac{1-c}{2c^2}+\frac{1}{2}-\frac{1}{2c}\bigg)(\log x)^{-1}, \end{align} $$

thanks to the estimate (2.6). Combining (2.9), (2.10) and (2.11), one sees that the right-hand side of (2.7) equals

(2.12) $$ \begin{align} \frac{1+o(1)}{\mathfrak{S}}\bigg(\frac{1}{2c}-\frac{1}{2}\bigg)(\log x)^{-1}. \end{align} $$

Taking (2.12) into (2.4), we immediately obtain

$$ \begin{align*} T^{\prime}_c(x)&\le (1+o(1))8\bigg(\frac{1}{c}-1\bigg)\frac{x}{\log x}. \end{align*} $$

Therefore, by Lemma 2.2,

$$ \begin{align*} T_c(x)=T^{\prime}_c(x)+O\bigg(\frac{x\log\log x}{(\log x)^2}\bigg)\le (1+o(1))8\bigg(\frac{1}{c}-1\bigg)\frac{x}{\log x}. \end{align*} $$

Our theorem now follows from the prime number theorem.

3. Remarks

Under the assumption of the Elliott–Halberstam conjecture, it is reasonable to predict that the exact value of c in Corollary 1.4 should be $e^{-1/2}=0.60653\ldots $ from (1.3) and the recursion formula (see, for example, [20, (7.6)]) for Dickman’s function:

$$ \begin{align*} \rho(v)=u-\int_{u}^{v}\frac{\rho(t-1)}{t}\,dt \quad (1\le u\le v). \end{align*} $$

It therefore seems to be of interest to improve, as far as possible, the numerical value of c in Corollary 1.4. We leave this as a challenge to readers.

Though we provided nontrivial upper bounds on $T_c(x)$ for $8/9\le c<1$ in Theorem 1.3, the extension of these bounds to $1/2\le c<1$ is an unsolved problem.